Page 143 - Electromagnetics

Then, since r = ˆ zz + ˆρρ, r = ˆ zz , and dl = dz ,wehave

∞

ρ l ˆ ρρ + ˆ z(z − z )

E(ρ) = 3/2 dz .
4π −∞ ρ + (z − z )
2

2
Carrying out the integration we find that E has only a ρ-component which varies only
with ρ:
ρ l
E(ρ) = ˆρ . (3.64)
2π ρ
The absolute potential referred to a radius ρ 0 can be found by computing the line integral
of E from ρ to ρ 0 :
ρ
ρ l dρ ρ l ρ 0
(ρ) =− = ln .
2π ρ 2π ρ
ρ 0
We may choose any reference point ρ 0 except ρ 0 = 0 or ρ 0 =∞. This choice is equivalent
to the addition of an arbitrary constant, hence we can also write

ρ l 1
(ρ) = ln + C. (3.65)
2π ρ
The potential for a general two-dimensional charge distribution in unbounded space is
by superposition

ρ T (ρ )

(ρ) = G(ρ|ρ ) dS , (3.66)

S T
where the Green’s function is the potential of a unit line source located at ρ :

1 ρ 0

G(ρ|ρ ) = ln . (3.67)
2π |ρ − ρ |

Here S T denotes the transverse (xy)plane, and ρ T denotes the two-dimensional charge
2
distribution (C/m )within that plane.
We note that the potential field (3.66)of a two-dimensional source decreases logarith-
mically with distance. Only the potential produced by a source of finite extent decreases
inversely with distance.
Dirichlet and Neumann Green’s functions. The unbounded space Green’s func-
tion may be inconvenient for expressing the potential in a region having internal surfaces.
In fact, (3.56)shows that to use this function we would be forced to specify both and its
normal derivative over all surfaces. This, of course, would exceed the actual requirements
for uniqueness.
Many functions can satisfy (3.52). For instance,
A B

G(r|r ) = + (3.68)

|r − r | |r − r i |
satisfies (3.52)if r i /∈ V . Evaluation of (3.55)with the Green’s function (3.68)repro-
duces the general formulation (3.56)since the Laplacian of the second term in (3.68)is
identically zero in V . In fact, we can add any function to the free-space Green’s function,
provided that the additional term obeys Laplace’s equation within V :
A
2
G(r|r ) = + F(r|r ), ∇ F(r|r ) = 0. (3.69)

|r − r |

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